[POJ1934] Trip】的更多相关文章

[题解]POJ1934 Trip (DP+记录方案) 题意: 传送门 刚开始我是这么设状态的(谁叫我DP没学好) \(dp(i,j)\)表示钦定选择\(i\)和\(j\)的LCS,然而你会发现这样钦定没什么用. 还不如当时初学者的时候的\(dp(i,j)\)表示考虑到\(i\)考虑到\(j\)的LCS...果然经典的是禁得起考验的... 考虑如何记录方案,第一个想法是直接暴力记录从哪转移的,但是这样显然不行.因为有很多重复的元素. 注意到题目保证本质不同的满足答案要求的串的个数是\(O(n)\)…
Trip Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3850   Accepted: 1030 Description Alice and Bob want to go on holiday. Each of them has planned a route, which is a list of cities to be visited in a given order. A route may contain a…
题目传送门:https://www.luogu.org/problemnew/show/SP33 题目大意:给出两个字符串,求其LCS(最长公共子序列)的长度与具体方案(相同的串算作同一方案).数据组数$\leq 10$,字符串长度$\leq 80$,方案数$\leq 1000$ 本来以为这是一道LCS水题,结果超级low的各种输出方案的方法TLE到怀疑人生 于是一个高大上的输出方法出现(借鉴于https://blog.csdn.net/gg_gogoing/article/details/41…
问题描述 Alice and Bob want to go on holiday. Each of them has planned a route, which is a list of cities to be visited in a given order. A route may contain a city more than once. As they want to travel together, they have to agree on a common route. No…
Luogu Description 爱丽丝和鲍伯想去度假,他们每个人都制定了一个参观城市的清单,该地区正好有26个城市,因此它们被编码为小写字母“a”到“z”.清单上可能重复出现某个城市.因为他们想一起旅行,他们必须走同样的路线.他们不想增加城市或改变城市顺序,所以只能从名单上删除一些城市.他们想尽可能多地参观城市.请你将他们两人的清单合成一个清单. Input 输入的第一行包含一个整数,表示测试用例数T(T<=10).每个测试用例有两行:第一行是爱丽丝的列表,第二行是鲍伯的列表.每个列表包含1…
LCS 模板,但要输出具体方案,这就很毒瘤了. 神奇的预处理:fa[i][j]表示在 \(a\) 串的前 \(i\) 个字符中,字母表第 \(j\) 个字母最晚出现的位置,fb[i][j]同理. 这样我们便可以递归找路径,对于两指针的位置的字母相同就向下递归,否则用预处理数组枚举相同位置再递归. 还要按字典序输出,开个set就行了. #include <bits/stdc++.h> using namespace std; const int N=85; char a[N],b[N],s[N]…
POJ2279 Mr. Young's Picture Permutations 题意 Language:Default Mr. Young's Picture Permutations Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2513 Accepted: 960 Description Mr. Young wishes to take a picture of his class. The students will…
题目意思: 有两个字符串(长度\(<=80\)),按字典序输出它们的最长公共子串的所有情况. 解析 最长公共子序列的长度应该都没问题了吧...有问题请自行百度 但关键是要求出每种情况,还要按字典序... 考虑到长度\(<=80\),我们可以用搜索+剪枝, 设两个串为\(a\),\(b\),长度为\(la\),\(lb\), \(f[i][j]\)表示\(a\)串中的\(1\)~\(i\)与\(b\)串中的$ 1\(~\)j$的最长公共子序列的长度. 先求出最长公共子序列的长度\(len\)=\…
Text I have just received a letter from my brother,Tim. He is in Australia. He has been there for six months. Tim is an engineer. He is working for a big firm and he has already visited a great number of different places in Australia. He has just bou…
Road Trip Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB Total submit users: 29, Accepted users: 29 Problem 12882 : No special judgement Problem description You are planning a road trip to visit your friends, each of whom live in…
http://poj.org/problem?id=1041 (题目链接) 题意 给出一张无向图,求字典序最小欧拉回路. Solution 这鬼畜的输入是什么心态啊mdzz,这里用vector储存边,便于边的排序.瞬间变成STL常数boy →_→. 细节 数组大小把握好. 代码 // poj1041 #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #incl…
A number of students are members of a club that travels annually to exotic locations. Their destinations in the past have included Indianapolis, Phoenix, Nashville, Philadelphia, San Jose, and Atlanta. This spring they are planning a trip to Eindhove…
Texas Trip Problem's Link:   http://poj.org/problem?id=3301 Mean: 给定n(n <= 30)个点,求出包含这些点的面积最小的正方形的面积. analyse: 首先要确定的是旋转的角度在0到180度之间即可,超过180度是和前面的相同的. 坐标轴旋转后,坐标变换为: X’ = x * cosa - y * sina; y’ = y * cosa + x * sina; Time complexity: O(n) Source code…
Problem C: The Trip Time Limit: 1 Sec  Memory Limit: 64 MBSubmit: 19  Solved: 3[Submit][Status][Web Board] Description The Trip A group of students are members of a club that travels annually to different locations. Their destinations in the past hav…
Ant Trip Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Ant Country consist of N towns.There are M roads connecting the towns. Ant Tony,together with his friends,wants to go through every part…
Sightseeing Trip Time limit: 0.5 secondMemory limit: 64 MB There is a travel agency in Adelton town on Zanzibar island. It has decided to offer its clients, besides many other attractions, sightseeing the town. To earn as much as possible from this a…
Mishka and trip 题意: 有n个城市,第i个城市与第i+1个城市相连,他们边的权值等于i的美丽度*i+1的美丽度,有k个首都城市,一个首都城市与每个城市都相连,求所有边的权值. 题解: 先把n个城市存下来,之后开一个标记数组,来标记k个首都(这题这块很巧妙,正因为开了标记首都的数组,所以才把O(N^2)的算法降到了O(N)) 把所有城市的美丽值都加起来,遍历首都,每次遍历完就sum-首都,这样,最后求环的剩下的边就好了,环的剩下的边就是i不是首都,i+1也不是首都,那么就i*i+1…
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2425 Hiking Trip Description Hiking in the mountains is seldom an easy task for most people, as it is extremely easy to get lost during the trip. Recently Green has decided to go on a hiking trip. Unfort…
题目链接:1484 - Alice and Bob's Trip 题意:BOB和ALICE这对狗男女在一颗树上走,BOB先走,BOB要尽量使得总路径权和大,ALICE要小,可是有个条件,就是路径权值总和必须在[L,R]之间,求终于这条路径的权值. 思路:树形dp,dp[u]表示在u结点的权值,往下dfs的时候顺带记录下到根节点的权值总和,然后假设dp[v] + w + sum 在[l,r]内,就是能够的,状态转移方程为 dp[u] = max{dp[v] + w }(bob) dp[u] = m…
Taxi Trip Time Winners' Interview: 3rd place, BlueTaxi This spring, Kaggle hosted two competitions with the ECML PKDD conference in Porto, Portugal. The competitions shared a dataset but focused on different problems. Taxi Trajectory asked participan…
#include<cstdio> #include<iostream> #include<string> #include<algorithm> #include<iterator> using namespace std; /* * A solution for "The Trip" problem. * UVa ID: 10137 */ #include <stdio.h> int main (int argc…
B. Mishka and trip time limit per test 1 second memory limit per test 256 megabytes input  standard input output  standard output  Little Mishka is a great traveller and she visited many countries. After thinking about where to travel this time, she…
Hiking in the mountains is seldom an easy task for most people, as it is extremely easy to get lost during the trip. Recently Green has decided to go on a hiking trip. Unfortunately, half way through the trip, he gets extremely tired and so needs to…
Qwerty78 Trip time limit per test 2 seconds memory limit per test 64 megabytes input standard input output standard output Qwerty78 is a well known programmer (He is a member of the ICPC WF winning team in 2015, a topcoder target and one of codeforce…
Description After a day trip with his friend Dick, Harry noticed a strange pattern of tiny holes in the door of his SUV. The local American Tire store sells fiberglass patching material only in square sheets. What is the smallest patch that Harry nee…
算一下路程和速度... #include <iostream> #include <cstdio> using namespace std; const double p=3.1415927; double dia,tim,dis,sped; int rev; int main() { ; while(cin>>dia>>rev>>tim,rev) { dia/=*; dis=dia*p*rev; sped=dis/tim**; printf(&…
题目链接 求给出的图是否存在欧拉回路并输出路径, 从1这个点开始, 输出时按边的升序输出. 将每个点的边排序一下就可以. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #in…
欧拉回路第一题TVT 本题的一个小技巧在于: [建立一个存放点与边关系的邻接矩阵] 1.先判断是否存在欧拉路径 无向图: 欧拉回路:连通 + 所有定点的度为偶数 欧拉路径:连通 + 除源点和终点外都为偶数 有向图: 欧拉回路:连通 + 所有点的入度 == 出度 欧拉路径:连通 + 源点 出度-入度=1 && 终点 入度 - 出度 = 1 && 其余点 入度 == 出度: 2.求欧拉路径 : step 1:选取起点(如果是点的度数全为偶数任意点为S如果有两个点的度数位奇数取一…
题意:给出n个包的大小,规定一个大包能装一个小包,问最少能装成几个包. 只要排序,然后取连续出现次数最多的数的那个次数.输出注意需要等距输出. 代码: /* * Author: illuz <iilluzen[at]gmail.com> * Blog: http://blog.csdn.net/hcbbt * File: uva11100.cpp * Lauguage: C/C++ * Create Date: 2013-09-03 16:25:50 * Descripton: uva1110…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3018 Ant Trip Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1658    Accepted Submission(s): 641 Problem Description Ant Country consist of N to…